Chapter 2, Problem 131CR

To determine

To graph: the function $f(x)$ and check for intercept and various asymptotes.

Answer to Problem 131CR

y-intercept is $2$

x-intercept is $\infty$

Vertical asymptote is at $x=-1$

Horizontal asymptote is at $y=0$

No holes

No slant asymptotes

Explanation of Solution

Given information:

  $f(x)=\frac{2}{{\left(x+1\right)}^2}$

Graph: Assuming the value of x to find $f(x)$ and plot the graph

  

  

Interpretation :

To determine y-intercept put $x=0$ and solve for $y$ or $f(x)$

  $f(x)=\frac{2}{{\left(x+1\right)}^2}=\frac{2}{{\left(0+1\right)}^2}=2$

To determine x-intercept put $y=0$ and solve for $x$

  $f(x)=\frac{2}{{\left(x+1\right)}^2}\Rightarrow 0=\frac{2}{{\left(x+1\right)}^2}\Rightarrow {\left(x+1\right)}^2=\frac{2}{0}\Rightarrow x=\infty$

To find the vertical asymptotes we have to solve the denominator by equating it equal to zero:

  ${\left(x+1\right)}^2=0\Rightarrow x+1=0\Rightarrow x=-1$

A horizontal asymptotes is defined when the degree of the denominator is greater than or equal to degree of the numerator.

Here the degree of denominator is greater than numerator therefore we calculate $\underset{x\to \infty }{\lim }f(x)$

  $\underset{x\to \infty }{\Rightarrow \lim }f(x)=\underset{x\to \infty }{\lim }\frac{2}{{\left(x+1\right)}^2}=0$

As $x$ gets bigger $f(x)$ gets nearer to zero

Therefore horizontal asymptote is at $y=0$

Slant asymptotes occur when the degree of denominator is lower than that of the numerator. since the function is having horizontal asymptotes and degree of denominator is greater than that of the numerator, therefore slant asymptote is not possible.

Now , the degree of the denominator is greater than the degree of the numerator therefore there will be no hole possible in the graph.